Question

A farmer has 20 yards of fencing to build a pen for her chickens. She decides to use a side of her barn as one side of the fenced-in area. What is the maximum area she can achieve?

Answer 1

Answer:

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.

Step-by-step explanation:

Given that farmer shall construct a rectangular fenced-in area and a side of the barn is one side of such area, the needed length of fencing is represent by the following perimeter equation ($p$), measured in square yards:

$p = 2\cdot l + w$

Where:

$l$ - Length of the rectangle, measured in yards.

$w$ - Width of the rectangule (side of the barn), measured in yards.

In addition, the equation of the fenced-in area ($A$) is:

$A = w\cdot l$

If $p = 20\,yd$, equation of area is now simplified as follows:

$A = (20\,yd - 2\cdot l)\cdot l$

$A = 20\cdot l - 2\cdot l^{2}$

The value of $l$ associated with the maximum area is obtained with the help of First and Second Derivative Tests. Firstly, first and second derivatives of the area function are determined:

$A' = 20 - 4\cdot l$

$A'' = -4$

Let equalize first equation to zero, second derivative indicates that critical value follows to an absolute maximum. Hence:

$20-4\cdot l = 0$

$l = 5\,yd$

The width of the rectangle is: ($p = 20\,yd$ and $l = 5\,yd$)

$w = p - 2\cdot l$

$w = 20\,yd - 2\cdot (5\,yd)$

$w = 10\,yd$

And finally, the maximum area she can achieve is:

$A = (5\,yd)\cdot (10\,yd)$

$A = 50\,yd^{2}$

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.